Project: Benchmark a Device Model

You cannot trust a quantum processor you have not characterized. Randomized benchmarking (RB) is the field's standard answer: it measures an average gate error that is robust to state-preparation and measurement (SPAM) errors, by averaging over random Clifford sequences. This capstone builds the fundamental RB primitive — a random sequence plus its exact recovery — whose survival probability is the data point a full RB experiment fits.

The RB protocol

For each sequence length $m$ :

Draw $m$ random Clifford gates $C_1, \dots, C_m$ .
Append the single recovery Clifford $C_{m+1} = (C_m \cdots C_1)^{-1}$ , so the ideal net operation is the identity.
Run, measure, and record the survival probability $P(|0\rangle)$ .
Repeat over many random sequences and average.

On a noiseless device every sequence returns the qubit to $|0\rangle$ , giving survival $1$ regardless of $m$ . On real hardware errors accumulate and the averaged survival decays.

The decay model

Averaging over the Clifford group depolarizes the errors, so the mean survival follows a clean exponential in the sequence length:

\bar{P}(m) = A\,p^{\,m} + B,

where $p$ is the depolarizing parameter and $A, B$ absorb SPAM. Fitting $p$ yields the average error per Clifford

r = \frac{(d - 1)(1 - p)}{d}, \qquad d = 2^n,

which is what gets quoted as a device's gate fidelity. Because the form of the decay is fixed and $A, B$ soak up state-prep and readout errors, the extracted $r$ isolates gate quality — RB's key advantage over naive process tomography.

A length-3 instance

Take the sequence $H, S, H$ . Its inverse, applied in reverse with each gate inverted, is $H, S^\dagger, H$ (since $H^{-1} = H$ and $S^{-1} = S^\dagger$ ). Concatenated, the circuit is the identity, so a perfect simulator returns survival $P(|0\rangle) = 1$ — the noise-free reference against which a noisy device's decay is measured.

Try it

Build the length-3 sequence and its recovery, then measure. The grader checks the output distribution: all probability on outcome $0$ , the ideal survival of $1$ .

Run your code to see the quantum state.